Question

p=(3^2000)+(3^-2000) and q=(3^2000)-(3^-2000) than value p^2-q^2 equalto=

Answer 1

$p=3^{2000}+3^{-2000}\\q=3^{2000}-3^{-2000}$

$p^2-q^2=(p+q)(p-q)$

$=[(3^{2000}+3^{-2000})+(3^{2000}-3^{-2000})]$ × $[(3^{2000}+3^{-2000})-(3^{2000}-3^{-2000})]$

$=[2*3^{2000}]*[2*3^{-2000}]\\ \\=4*[3^{2000}*3^{-2000}]$

$=4*[3^{2000}* \frac{1}{3^{2000}} ]\\ \\= \boxed4$

Answer 2

Answer:

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